Maximum possible value

For $a, b, c > 0$ , we can apply the Cauchy-Schwarz inequality .

$\begin{aligned} \left(\sqrt{2a+b} + \sqrt{2b+c} + \sqrt{2c+a} \right)^2 & \le \left(1^2+1^2+1^2\right)\left((\sqrt{2a+b})^2 + (\sqrt{2b+c})^2 + (\sqrt{2c+a})^2 \right) \\ & \le 3 \left(2a+b + 2b+c + 2c+a \right) = 9(a+b+c) = 3600 \\ \implies \sqrt{2a+b} + \sqrt{2b+c} + \sqrt{2c+a} & \le \boxed{60} \end{aligned}$

Due to symmetry, a=b=c.

By using that we get 60,which is the maximum

value in the option.

Therefore, 60 is the answer.